# Equation of a Plane in the Normal Form

The equation of a plane in the normal form can be determined if normal to the plane as well as the distance of the plane from the origin is known.

### Equation of a Plane in the Normal Form

The vector form of the equation of a plane in normal form is given by:

Where \(\vec{r}\) is the position vector of a point in the plane, *n* is the unit normal vector along the normal joining the origin to the plane and *d* is the perpendicular distance of the plane from the origin.

Let us say that P (x, y, z) is any point on the plane and O is the origin. Then, we have,

Now we take the direction cosines of \(\hat{n}\) as *l*, *m* and *n*. So we have,

From the equation \(\vec{r}.\hat{n}\) = d we get

Thus, the Cartesian form of the equation of a plane in normal form is given by:

*lx + my + nz = d*

### Example

Let us take up an example to understand the equation of a plane in the normal form.

Problem: A plane is at a distance of \(\frac{9}{\sqrt{38}}\) from the origin O. From the origin, its normal vector is given by 5\(\hat{i}\) + 3\(\hat{j}\) – 2\(\hat{k}\).

What is the vector equation for the plane?

Solution: Let the normal vector be:

We now find the unit vector for the normal vector. It can be given by:

So, the required equation of the plane can be given by substituting it in the vector equation: